# Couple Questions im stuck on.

• Nov 11th 2006, 07:09 PM
rusty26
Couple Questions im stuck on.
1) Let F: X -> Y be a function. Justify your answer to each question below by giving either a proof or a counter example.

a) Does f(A - B) = f(A) - f(B), for all A,B subsets of X?

b) Does f(inverse)(C - D) = f(inverse)(C) - f(inverse)(D), for all C,D subset of Y.

Just a not, when I say A - B that is the difference between the sets.

I think I know somewhat how to approach this> I believe I have to prove through the use of set images and that such, but I am stuck. Any help would be great.
• Nov 11th 2006, 07:30 PM
ThePerfectHacker
Quote:

Originally Posted by rusty26
1) Let F: X -> Y be a function. Justify your answer to each question below by giving either a proof or a counter example.

a) Does f(A - B) = f(A) - f(B), for all A,B subsets of X?

Write it out symbolically.

$f[A]=\{f(x)|x\in A\}$
$f[ B ]=\{f(x)|x\in B\}$
$A-B=\{x\in A, x\not \in B\}$
$f[A-B]=\{f(x)|x\in A,x\not \in B\}$
$f[A]-f[ B ]=\{x|x\in f[A], x\not \in f[ B ]\}=\{f(x)|x\in A, x\not \in B\}$
So it seem to me that,
$f[A-B]=f[A]-f[ B ]$
• Nov 12th 2006, 12:19 AM
rusty26
Ahh, thanks alot.

How would I apply that to the inverse, would it be similar but just using the complement of sets?
• Nov 12th 2006, 06:03 AM
Plato
Unfortunately it is the case that $f\left[ {A\backslash B} \right] \not= f\left[ A \right]\backslash f\left[ B \right]$. Consider the function $f:\left\{ {1,2,3,4} \right\} \mapsto \left\{ {a,b,c} \right\}\quad f = \left\{ {\left( {1,a} \right),\left( {2,b} \right),\left( {3,b} \right),\left( {4,c} \right)} \right\}$
Then look at $\begin{array}{l}
f\left[ {\left\{ {1,2,3} \right\}\backslash \{ 3,4\} } \right] = f\left[ {\left\{ {1,2} \right\}} \right] = \left\{ {a,b} \right\} \\
f\left[ {\left\{ {1,2,3} \right\}} \right]\backslash f\left[ {\left\{ {3,4} \right\}} \right] = \left\{ {a,b} \right\}\backslash \left\{ {b,c} \right\} = \left\{ a \right\} \\
\end{array}.$
• Nov 12th 2006, 06:17 AM
Plato
To prove $\overleftarrow f \left[ {K\backslash L} \right] = \overleftarrow f \left[ K \right]\backslash \overleftarrow f \left[ L \right]$ is true consider this.
$
\begin{array}{rcl}
x \in \overleftarrow f \left[ {K\backslash L} \right] & \Leftrightarrow & f(x) \in K\backslash L \\
& \Leftrightarrow & f(x) \in K \wedge f(x) \notin L \\
& \Leftrightarrow & x \in \overleftarrow f \left[ K \right] \wedge x \notin \overleftarrow f \left[ L \right] \\
& \Leftrightarrow & x \in \left( {\overleftarrow f \left[ K \right]\backslash \overleftarrow f \left[ L \right]} \right) \\
\end{array}
$
• Nov 12th 2006, 06:52 AM
ThePerfectHacker
Does,
A/B mean A-B? If so then,
And what did I make a mistake upon?
• Nov 12th 2006, 07:38 AM
Plato
There is a quantification error in line 4 of your work.
Look carefully at the definition: $x \in \overrightarrow f \left( {A\backslash B} \right) \Leftrightarrow \left( {\exists t \in A\backslash B} \right)\left[ {f(t) = x} \right].$
There is an existential quantifier in the definition of the image function.
In the counterexample you can see how that works. In the preimage function the quantification is universal. In fact, image of an intersection is only a subset of the intersection of the images. That is the only case where it fails. Recall that set difference is really an intersection.

And yes A\B is the preferred notation. Even in TeX $\setminus$ is ‘setminus’.
• Nov 12th 2006, 08:37 AM
rusty26
ok, so part A all i need to do is give proof by the counterexample that Plato has shown, however, does that mean that for part B it would also not be true?
• Nov 12th 2006, 10:08 AM
Plato
Quote:

Originally Posted by rusty26
ok, so part A all i need to do is give proof by the counterexample that Plato has shown, however, does that mean that for part B it would also not be true?

Yes part B is correct. The proof is in my second reply.
The symbol ${\overleftarrow f \left[ L \right]}$ is the inverse or premiage of L.